Permutation Inclusion Solution

Problem Statement

Given a key phrase and a longer intercepted message, return true if any rearrangement of the key phrase exists as a continuous segment within the message.

Example 1

Input

keyPhrase = 'xyz', message = 'abcxyzdef'

Output

false

Explanation: Step 1: Initialize two pointers, one for the key phrase and one for the message. Step 2: Compare the characters at the current positions of the key phrase and the message. If they match, move both pointers forward. If they don't match, move the pointer for the key phrase forward. Step 3: If the key phrase pointer reaches the end of the key phrase, it means we have found a permutation of the key phrase in the message, so return true. Otherwise, return false.

Example 2

Input

keyPhrase = 'xyz', message = 'xyzabcxyzdef'

Output

true

Explanation: Step 1: Initialize two pointers, one for the key phrase and one for the message. Step 2: Compare the characters at the current positions of the key phrase and the message. If they match, move both pointers forward. If they don't match, move the pointer for the key phrase forward. Step 3: If the key phrase pointer reaches the end of the key phrase, it means we have found a permutation of the key phrase in the message, so return true. Otherwise, return false.

Constraints

1 <= s1.length, s2.length <= 10^4
s1 and s2 consist of lowercase English letters.

Problem Statement

Input

Output

Input

Output

Constraints

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Optimized Approach

Step-by-Step Dry Run

Tested Solutions